How to create a new tuple type from an old one and a type in boost?

Question

I have a tuple type. I want to add a element type in it to get a new tuple type. I can do it like

decltype tuple_cat(MyTuple, std::tuple())
         


        

Answer 1

I assume you want all this in compile time.

Here is the general explanation: concatening tuples is similar to concatening lists or arrays, is that the algorithm is the same. Here, given tuples a and b, I choosed to move the last element of a to the beginning of b, and repeat until a is empty.

First: base structures. The following structure keeps a parameter pack. It can be anything, for example a tuple:

template
struct pack
{
    static const unsigned int size = sizeof...(T);
};

Note that the size of the pack is stored inside it. It is not mandatory, but it is convenient for the explanation. Boost uses the struct boost::tuples::length::value (which is more verbose).

To access an element at i-th position, we use a structure similar to boost::tuples::element:

// Get i-th element of parameter pack
// AKA 'implementation'
// Principle: the element i is the first element of the sub-array starting at indice i-1
template
struct element_at : public element_at
{
};

template
struct element_at<0, F, T...>
{
    typedef F type;
};

// Get i-th element of pack
// AKA 'interface' for the 'pack' structure
template
struct element
{
};

template
struct element>
{
    typedef typename element_at::type type;
};

Now, we must use a low-level operation which is adding one element to a side of a pack (adding at left or at right). Here adding at left is choosed, but it is not the only choice:

// Concat at left (only for structure 'pack')
template
struct tuple_concat_left
{
};

template
struct tuple_concat_left>
{
    typedef pack type;
};

For templates, a is not changed, and instead we use an indice to know what element to add. The inheritance define a 'type' typedef which is the concatenation of all indices after n and the other tuple (not including n, and in order). We just have to concatenate at left the element at indice n.

// Concat 2 tuples
template
struct tuple_concat : public tuple_concat
{
    typedef typename tuple_concat_left<
        typename element::type,
        typename tuple_concat::type
    >::type type;
};

template
struct tuple_concat
{
    typedef b type;
};

And that's it! Live example here.

Now, for tuple specifics: you noticed I didn't used boost::tuple nor std::tuple. That is because a lot of implementations of boost tuplesdo not have access to variadic templates, so a fixed number of template parameters is used (they default to boost::tuples::null_type). Putting this directly with variadic templates is a headache, thus the need to have another abstraction.

I also assumed that you can use C++11 (with the decltype in your question). Concatening 2 tuples in C++03 is possible, but more repetitive and boring.

You can convert a pack to a tuple really easily: just change the pack definition to:

template
struct pack
{
    static const unsigned int size = sizeof...(T);
    typedef boost::tuple to_tuple; // < convert this pack to a boost::tuple
};